From a645391babeeb841d1949070d0c4bba7b48cb850 Mon Sep 17 00:00:00 2001
From: 8057a7ae58e587eb7d45460ceced4594
 <8057a7ae58e587eb7d45460ceced4594@app-learninglab.inria.fr>
Date: Thu, 23 Sep 2021 17:49:08 +0000
Subject: [PATCH] minor changes

---
 module2/exo1/toy_notebook_en.ipynb | 36 +++++++-----------------------
 1 file changed, 8 insertions(+), 28 deletions(-)

diff --git a/module2/exo1/toy_notebook_en.ipynb b/module2/exo1/toy_notebook_en.ipynb
index 03952de..90c97c5 100644
--- a/module2/exo1/toy_notebook_en.ipynb
+++ b/module2/exo1/toy_notebook_en.ipynb
@@ -4,20 +4,10 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "# On the computation of $\\pi$"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Asking the maths library"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
+    "# On the computation of $\\pi$\n",
+    "\n",
+    "## Asking the maths library\n",
+    "\n",
     "My computer tells me that $\\pi$ is *approximatively*"
    ]
   },
@@ -82,13 +72,8 @@
    "cell_type": "markdown",
    "metadata": {},
    "source": [
-    "## Using a surface fraction argument"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
+    "## Using a surface fraction argument\n",
+    "\n",
     "A method that is easier to understand and does not make use of thesin function is based on thefact that if $X \\sim U(0,1)$ and $Y \\sim U(0,1)$, then $P[X^2 + Y^2 \\leq 1] = \\frac{\\pi}{4}$ (see [\"Monte Carlo method\" on Wikipedia](https://en.wikipedia.org/wiki/Monte_Carlo_method). The following code uses this approach:"
    ]
   },
@@ -113,12 +98,14 @@
    "source": [
     "%matplotlib inline\n",
     "import matplotlib.pyplot as plt\n",
+    "\n",
     "np.random.seed(seed=42)\n",
     "N=1000\n",
     "x=np.random.uniform(size=N, low=0, high=1)\n",
     "y=np.random.uniform(size=N, low=0, high=1)\n",
     "accept=(x*x+y*y)<=1\n",
     "reject=np.logical_not(accept)\n",
+    "\n",
     "fig, ax=plt.subplots(1)\n",
     "ax.scatter(x[accept], y[accept], c='b', alpha=0.2, edgecolor=None)\n",
     "ax.scatter(x[reject], y[reject], c='r', alpha=0.2, edgecolor=None)\n",
@@ -151,13 +138,6 @@
    "source": [
     "4*np.mean(accept)"
    ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": []
   }
  ],
  "metadata": {
-- 
2.18.1